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# An ionic compound has a unit cell consisting of A ions at the corners of a cube and B ions on the centres of the faces. If side length is 10$A^{\large\circ}$, what is the density of the solid assuming atomic weights of A and B as 50 and 100 gms respectively.

$\begin{array}{1 1} 1.082 \times 10^6 \; gm^{-3} \\5.812\; g\;cm^{-3} \\ 6.324 \times 10^6 \; gm^{-3} \\ 4.346 \; g\;cm^{-3} \end{array}$

Answer: $5.812\; g\;cm^{-3}$
Number of A ions per unit cell = $\large\frac{1}{8}$$\times 8 = 1 Number of B ions per unit cell = \large\frac{1}{2}$$\times 6 = 3$
$\Rightarrow$ the empirical formula is $AB_3$
$\Rightarrow$ the mass of the compound $= 50 + 3 \times 100 = 350\; gm$
The density is given by the formula $\large\frac{z\;M}{a^3\;N_A}$
Avagadro's number $N_A = 6.022 \times 10^{23} /\; mol$
Given $z = 1,\; M = 350 \; gm, \; a = 10^{\circ}A = 10 \times 10^{-8}cm$, we can calculate density as follows:
$\Rightarrow$ Density $\rho = \large\frac{ 1 \times 350}{ (10 \times 10^{-8})^3 \times 6.022 \times 10^{23}}$$= 5.812\; g\;cm^{-3}$
edited Jul 15, 2014